Proof that π is irrational
Lisa Nicklasson
Tomas Berggren
We will do a proof by contradiction. Assume that π is rational, that is,
π = ab for some integers a and b. For some positive integer n define two
functions f (x) and F (x) as
f (x) =
xn (a − bx)n
,
n!
and
F (x) = f (x) − f (2) (x) + f (4) (x) − · · · + (−1)n f (2n) .
We start by noting some properties and relations between these functions.
The polynomial f (x) is of degree 2n. This implies that f (k) (x) is identically
zero for k greater than 2n and we get the relation
F 00 (x) + F (x) = (f (2) (x) − f (4) (x) − · · · + (−1)n−1 f (2n) )+
+ (f (x) − f (2) (x) + f (4) (x) − · · · + (−1)n f (2n) )
= f (x).
Also note that the lowest power that occur in f (x) is of degree n. Hence
f (k) (0) is zero when k is less than n, and f (k) (0) is an integer when n ≤ k ≤ 2n.
By this and the relation
f (a/b − x) =
(a/b − x)n (a − b(a/b − x))n
=
n!
1
(a
bn
− bx)n (bx)n
= f (x),
n!
we can conclude that also f (k) (π) is an integer. Therefore F (0) and F (π) are
integers.
By the product rule we get
d
(F 0 (x) sin(x) − F (x) cos(x)) = F 00 (x) sin(x) + F (x) sin(x) = f (x) sin(x)
dx
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and by the Fundamental Theorem of Calculus
Z π
f (x) sin(x)dx =F 0 (π) sin(π) − F (π) cos(π) − (F 0 (0) sin(0) − F (0) cos(0))
0
=F (0) − F (π)
which is an integer. From the definition of f (x) we get that
0 < f (x) sin(x) <
for 0 < x < π, hence
Z
0<
π
Z
f (x) sin(x)dx <
0
0
π
π n an
,
n!
π n an
(πa)n
dx = π
n!
n!
for all n ∈ N. But the right hand side converges to zero as n goes to ∞.
Then
Z π
f (x) sin(x)dx
0
can be made arbitrarily small, by choosing n large enough. But this contradicts the number being an integer, and hence we have proved π is irrational.
References
[1] Ivan Niven
A simple proof that π is irrational
Bulletin of the American Mathematical Society, Volume 53, nr. 6 (1947)
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